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How is temperature defined for solids?

  1. Jan 25, 2012 #1
    In my physics text book, temperature is defined as a measure of the average KE of random translational motion of particles. It also states that temperature does not account for the KE from molecular vibration or molecular rotation. In an earlier chapter, my text describes the motion of atoms in solids as vibrational motion - I take it that these atoms do not have translational motion. If this is the case, how is temperature measured for solids? Temperature would be the average KE of what?
     
  2. jcsd
  3. Jan 25, 2012 #2
    For a classical ideal gas, K/n = 0.5kT, where K is the kinetic energy of the particle(s), n is the number of degrees of freedom (the different ways the particle(s) can move in space), k is the Boltzmann's constant and T is the temperature.

    A monoatomic gas particle possesses 3 degrees of freedom:
    1. Its motion can be resolved along the x axis.
    2. Its motion can be resolved along the y axis.
    3. Its motion can be resolved along the z axis.
    Therefore, according to the above formula, the temperature of the monoatomic gas comes from the translational motion of the particles.

    A diatomic gas molecule possesses 6 degrees of freedom:
    1. The motion of its centre of mass can be resolved along the x axis.
    2. The motion of its centre of mass can be resolved along the y axis.
    3. The motion of its centre of mass can be resolved along the z axis.
    4. The molecule can vibrate along the axis joining the centres of the atoms.
    5. The molecule can rotate clockwise about the axis perpendicular to the midpoint of the line joining the atoms.
    5. The molecule can rotate anticlockwise about the axis perpendicular to the midpoint of the line joining the atoms.
    Therefore, according to the above formula, the temperature of the diaatomic gas comes not only from the translational motion of the centre of mass of the molecule, but also from the rotation and vibration of the molecule as a whole.

    The analysis can be continued for molecules of three or more atoms.

    Apparently, the declaration in the textbook assumes that the particles considered are monoatomic and that these are the constituents of an ideal gas (that is, the gas particles do not interact via forces).

    In an ideal gas, the interatomic and/or intermolecular forces are neglected, but in a solid, these cannot be ignored. The assumption of an ideal gas is one of the principle postulates in the derivation of the kinetic theory of gases (and the above relation for the temperature). Therefore, this relation cannot be used to understand temperature in solids. However, temperature is still a measure of the vibrational kinetic energy of the constituents of the solid.
     
  4. Jan 26, 2012 #3
    what about a monatomic liquid?
     
  5. Jan 26, 2012 #4
    At room temperature, this enumeration is not quite right, I think. There are FIVE essentially continuous quadratic degrees of freedom for a diatomic molecules. Your first three are correct (translation in x, y, z) but the extra two are rotational degrees of freedom, not clockwise and anticlockwise about a single perpendicular axis, but rotations about the TWO independent axes that are perpendicular to the line joining the two atoms (e.g. the y and z axes if the atoms are at +/-a along the x axis). Diatomic molecules tend to be very stiff in the vibrational mode so this degree of freedom is "frozen" at room temperature. That is why, according to the usual rule, the ratio of specific heats cp/cv = (5+2)/5 = 1.40.

    I disagree with the physics text that the OP quotes, if the quote is accurate. In particular "temperature does not account for the KE from molecular vibration or molecular rotation" is severely misleading, since energy is clearly absorbed by those modes when heat is added to a diatomic gas -- that's why the cv is different between gases of different structure.

    To answer the original question, in a crystalline solid at a high enough temperature, there are six quadratic degrees of freedom per atom in the crystal: 3 for translational KE, 3 for translational PE. Each atom accounts T/2 in the internal energy, so for one mole of atoms the specific heat is 3R (where R is the universal gas constant.) This is known as the law of Dulong and Petit.
     
  6. Jan 26, 2012 #5
    Thank you for your answers! bbbeard, you refer to the crystalline solid as having translational KE - So I take it the vibrational motion of the atom/molecule (the back and forth motion) is what accounts for the translational motion, and the "molecular vibration" is something completely different (as described above). Is that what I can take from this?
    Also, the text that I read about molecular vibration and molecular rotation not contributing to temperature, I understood this to mean that heat energy was absorbed, but did not result in an immediate temperature change. Instead, this energy accounted for a higher specific heat. Is that accurate?
     
  7. Jan 26, 2012 #6

    Vanadium 50

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    If you are going to define temperature via the properties of an ideal gas, a solid is at temperature T if it would be in thermal equilibrium with a gas at temperature T.
     
  8. Jan 27, 2012 #7
    Yes, the three translational degrees of freedom I'm talking about for a crystalline solid are the oscillation of the atom about its equilibrium site. This is analogous to the oscillation of the atoms in a covalently bonded molecule, except that in the latter the atoms are typically bound more tightly (though this depends on the molecule, obviously). It is the ratio of the vibrational energy level E=hbar*ω to the temperature T that determines whether that degree of freedom is "frozen" or not.

    When heat transfer changes the internal energy of a substance, if the substance is allowed to come to equilibrium, the energy is distributed among the various internal degrees of freedom such as translational and rotational degrees of freedom. (Classically, each degree of freedom soaks up the same amount of energy, though this is not true quantum mechanically. This is called "the principle of equipartition of energy".) That's why I think it is incorrect to state that temperature does not "account" for these additional degrees of freedom. Those degrees of freedom absorb energy -- that is why it takes more energy to raise the temperature a given amount when the substance has more degrees of freedom. Does that make sense?

    BBB
     
  9. Jan 27, 2012 #8
    OK - But one more clarification. If we could transfer heat so that it would be distributed only to the rotational degrees of freedom (which I understand would be impossible but please indulge me for one second), would an object already at thermal equilibrium remain so? If yes, then the object that absorbed the energy would, I believe, be more resistant to change its temperature. Accurate?
     
    Last edited: Jan 27, 2012
  10. Jan 27, 2012 #9
    I don't understand your hypothetical. Say you have a diatomic gas that is in thermal equilibrium with heat bath at a temperature T, and that the internal energy is distributed among the translational and rotational degrees of freedom as we have been describing. Then you want to energize just the rotational degrees of freedom, but suppress the equilibration processes that would normally redistribute that extra energy to the translational degrees of freedom? Well, I suppose it's your rules, so if you posit a way to prevent the energy from "leaking" into the translational degrees of freedom, then you can also posit a way to prevent that energy from leaking to the heat bath. But I think the energy leaks in a "classical" way -- imagine the molecule as a "dumbbell" hurtling through space. As it tumbles, it bumps into other molecules, transferring energy to both rotational and translational modes of the other molecules. For that matter, the tumbling molecule knocks into the walls of whatever separates it from the heat bath, and if the rotation is more energetic than usual, it will deposit energy into the heat bath until the extra rotational energy has equilibrated. I would say in that case that it is not in thermal equilibrium until the rotational modes have dumped their excess energy.

    But, like I said, it's your hypothetical. Whether or not the roto-excited gas is still in equilibrium depends on what you specify as the extra-physical means to suppress equilibration.
     
  11. Jan 28, 2012 #10
    In my earlier post regarding my hypothetical, I meant to say molecular vibrational degrees of freedom instead of rotational degrees of freedom, but I can see from your explanation above, it would come to the same result - average translational KE would ultimately increase, even if we could somehow distribute heat energy solely to the vibrational degrees of freedom. So in conclusion, temperature in a solid would be defined as the measure of the average KE of random translational motion of particles (the vibrations or wiggle motion of molecules in a solid for example).

    In addition, it is inaccurate to say that temperature does not account for molecular vibration (or molecular rotation)- because the molecular vibration can contribute to the average translational KE (vibrational movement in solids) of the molecules.

    I thank you and appreciate your time and patience in dealing with my questions.
     
  12. Jan 28, 2012 #11
    The definition of thermodynamic temperature for any system is

    [tex] T \equiv \left( \frac{\partial S}{\partial U} \right)^{-1} [/tex]

    where [itex]S[/itex] is entropy and [itex] U [/itex] internal energy.

    Applying this definition in kinetic theory gives the concept of kinetic temperature as average KE of translational motion of particles.
     
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